Question: Simplify the following expression: $y = \dfrac{-8x^2+31x- 21}{-8x + 7}$
Explanation: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-8)}{(-21)} &=& 168 \\ {a} + {b} &=& &=& {31} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $168$ and add them together. The factors that add up to ${31}$ will be your ${a}$ and ${b}$ When ${a}$ is ${7}$ and ${b}$ is ${24}$ $ \begin{eqnarray} {ab} &=& ({7})({24}) &=& 168 \\ {a} + {b} &=& {7} + {24} &=& 31 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-8}x^2 +{7}x) + ({24}x {-21}) $ Factor out the common factors: $ x(-8x + 7) - 3(-8x + 7)$ Now factor out $(-8x + 7)$ $ (-8x + 7)(x - 3)$ The original expression can therefore be written: $ \dfrac{(-8x + 7)(x - 3)}{-8x + 7}$ We are dividing by $-8x + 7$ , so $-8x + 7 \neq 0$ Therefore, $x \neq \frac{7}{8}$ This leaves us with $x - 3; x \neq \frac{7}{8}$.